Question

What are the coordinates of the focus of the parabola

Answer 1

(14, -8)

Explanation:

Given the function:

`y^2+16y+4x+4=0`

We want to find the coordinates of the focus of the parabola.

A parabola is the locus of points such that the distance to the focus equals the distance to the directrix. For a right-left facing parabola with vertex at (h, k), and a focal length |p|, the standard equation is:

`4p\left(x-h\right)=\left(y-k\right)^2`

First, we rewrite the given equation in the standard equation form:

`\begin{gathered} -4x-4=y^2+16y \\ -4x-4+64=y^2+16y+64 \\ -4x+60=(y+8)^2 \\ -4(x-15)=(y+8)^2 \\ \implies4\left(-1\right)\left(x-15\right)=\left(y-\left(-8\right)\right)^2 \\ \implies\left(h,k\right)=\left(15,-8\right),p=-1 \end{gathered}`

Next, our parabola is symmetric around the x-axis and so the focus lies at a distance p from the center (15,-8) along the x-axis. Thus, the coordinates of the focus is:

`(15+p,-8)=(15+(-1),-8)=(14,-8)`

The focus is at (14, -8).